The $n\text{th}$ partial sum of the series $\sum\limits_{n=1}^{\infty }{{{a}_{n}}}$ is given by ${{S}_{n}}=\frac{n^2-3}{n^3+4}$. $\sum\limits_{n=1}^{6}{{{a}_{n}}}=$
Answer: $\sum\limits_{n=1}^{6}{{{a}_{n}}}~~~$ is the same as ${{S}_{6}}\,$. ${{S}_{6}}=\frac{6^2 -3}{6^3 +4}=\frac{33}{220}=\frac{3}{20}$